Problem 53 - College Physics

Question

\(\bullet\) Exposing photographic film. The light-sensitive com- pound on most photographic films is silver bromide (AgBr). A film is "exposed" when the light energy absorbed dissociates this molecule into its atoms. (The actual process is more complex, but the quantitative result does not differ greatly.) The energy of dissociation of AgBr is \(1.00 \times 10^{5} \mathrm{J} / \mathrm{mol}\) . For a photon that is just able to dissociate a molecule of silver bromide, find (a) the photon's energy in electron volts, (b) the wavelength of the photon, and (c) the frequency of the photon. (d) Light from a firefly can expose photographic film, but the radiation from an FM station broadcasting \(50,000 \mathrm{W}\) at 100 \(\mathrm{MHz}\) cannot. Explain why this is so, basing your answer on the energy of the photons involved.

Step-by-Step Solution

Answer

Energy of the photon is 1.04 eV, wavelength is approximately 1200 nm, and frequency is 2.50 × 10¹⁴ Hz. Firefly light has higher energy photons compared to FM waves.

Step 1: Calculate Energy per Molecule

The energy of dissociation given is for one mole of silver bromide. To find the energy per molecule, use Avogadro's number, which is \(6.022 \times 10^{23} \text{ mol}^{-1}\). Divide the total energy by Avogadro's number:\[ E_{molecule} = \frac{1.00 \times 10^{5} \text{ J/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 1.66 \times 10^{-19} \text{ J/molecule}. \]

Step 2: Convert Energy from Joules to Electron Volts

Convert the energy per molecule from joules to electron volts using the conversion factor \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\):\[ E_{eV} = \frac{1.66 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 1.04 \text{ eV}. \]

Step 3: Calculate Wavelength of Photon

Use the Planck's relation that connects energy with wavelength: \(E = \frac{hc}{\lambda}\).Given \(E = 1.66 \times 10^{-19} \text{ J}\), Planck's constant \(h = 6.626 \times 10^{-34} \text{ Js}\), and speed of light \(c = 3.00 \times 10^{8} \text{ m/s}\), solve for wavelength \(\lambda\):\[ \lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^{8} \text{ m/s})}{1.66 \times 10^{-19} \text{ J}} \approx 1.20 \times 10^{-6} \text{ m},\] which is approximately 1200 nm.

Step 4: Calculate Frequency of Photon

Frequency \(u\) can be found using the speed of light and wavelength: \(c = \lambda u\). Using \(\lambda = 1.20 \times 10^{-6} \text{ m}\), solve for \(u\):\[ u = \frac{c}{\lambda} = \frac{3.00 \times 10^{8} \text{ m/s}}{1.20 \times 10^{-6} \text{ m}} \approx 2.50 \times 10^{14} \text{ Hz}. \]

Step 5: Explain Light from Firefly vs. FM Station

Light from a firefly has a higher frequency and therefore higher energy photon compared to FM radio waves. FM radio at 100 MHz has much lower frequency (and thus energy) than that required to break the AgBr bond. Hence, firefly light can expose the film but not radio waves.

Key Concepts

Silver Bromide Dissociation

Photographic films commonly contain silver bromide (AgBr) as their light-sensitive component. The process of exposing the film involves the dissociation of AgBr molecules into silver and bromine atoms due to absorption of energy from light photons. This dissociation requires a specific amount of energy, termed as the dissociation energy, which is crucial in this process.
The dissociation energy of AgBr is provided in terms of moles, specifically, as \(1.00 \times 10^{5} \text{ J/mol}\). To find how much energy is required to break just one molecule, we must divide the energy given for a mole by Avogadro's number. Since Avogadro's number is \(6.022 \times 10^{23} \text{ molecules/mol}\), this division calculates the energy necessary for dissociating a single molecule of AgBr, resulting in \(1.66 \times 10^{-19} \text{ J/molecule}\).Understanding the energy required for this dissociation is the basis for calculating the related properties of the corresponding photon, like its energy, wavelength, and frequency.

Photon Wavelength Calculation

To understand the effect of photon energy on silver bromide dissociation, we calculate the wavelength of a photon that holds enough energy to dissociate a molecule of AgBr. The relationship between energy and wavelength for a photon is provided by Planck’s equation, \(E = \frac{hc}{\lambda}\).Here, \(h\) is Planck’s constant \((6.626 \times 10^{-34} \text{ Js})\), \(c\) is the speed of light \((3.00 \times 10^{8} \text{ m/s})\), and \(E\) is the photon energy \( (1.66 \times 10^{-19} \text{ J})\). Plugging these values into the equation allows us to solve for the wavelength \(\lambda\) in meters:\[\lambda = \frac{hc}{E} = \frac{(6.626 \times 10^{-34} \text{ Js})(3.00 \times 10^{8} \text{ m/s})}{1.66 \times 10^{-19} \text{ J}} \approx 1.20 \times 10^{-6} \text{ m}\]Converting meters to nanometers (since 1 m = 1,000,000,000 nm), we find the wavelength to be approximately 1200 nm. This value indicates the size of the photon needed to dissociate one molecule of AgBr, which helps us understand the interaction of light with photo-sensitive materials.

Frequency Calculation

The frequency of a photon is yet another property that plays a vital role in the light sensitivity of silver bromide. The frequency (\(u\)) of a photon can be determined if its wavelength is known, using the equation \(c = \lambda u\). Here:

• \(c\) is the speed of light \((3.00 \times 10^{8} \text{ m/s})\)
• \(\lambda\) is the wavelength \((1.20 \times 10^{-6} \text{ m})\)

Rearranging the equation to solve for frequency:\[u = \frac{c}{\lambda} = \frac{3.00 \times 10^{8} \text{ m/s}}{1.20 \times 10^{-6} \text{ m}} \approx 2.50 \times 10^{14} \text{ Hz}\]Thus, the photon has a frequency of approximately \(2.50 \times 10^{14} \text{ Hz}\). This high frequency is necessary for the energy needed to excite and split the AgBr compound, showing why certain types of radiation can make a photographic film sensitive to light.

Energy Conversion to Electron Volts

When dealing with the energy of photons, it is often convenient to convert the energy from joules to electron volts (eV). This conversion makes it easier to relate the energy of photons with other processes in atomic and molecular physics. The conversion factor to switch from joules to electron volts is \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).Using the photon energy calculated in joules(\(1.66 \times 10^{-19} \text{ J}\)):\[E_{eV} = \frac{1.66 \times 10^{-19} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 1.04 \text{ eV}\]This shows that a photon with an energy of about 1.04 electron volts can dissociate a silver bromide molecule. Thus, by converting the energy into more applicable units, we can easily understand the interaction of various types of light with AgBr and why only certain energies suffice to expose photographic film. Additionally, this explains why visible light can cause a reaction, but lower energy radio waves often do not.